Korzystamy ze wzorów:

${\left(a\right)}^{-n}={\left(\frac{1}{a}\right)}^n$

$\frac{b^m}{b^n}=b^{m-n}$

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${\left(\frac{2}{3}\right)}^{-3}={\left(\frac{3}{2}\right)}^3=\frac{27}{8}$ ${\left(\frac{2}{3}\right)}^{-3}={\left(\frac{3}{2}\right)}^3=\frac{27}{8}$

${\left(\frac{2}{3}\right)}^{-1}={\left(\frac{3}{2}\right)}^1=\frac{3}{2}$

${\left(\frac{3}{2}\right)}^2=\frac{9}{4}$

${\left(\frac{2}{3}\right)}^3=\frac{8}{27}$

${\left(\frac{3}{2}\right)}^{-3}={\left(\frac{2}{3}\right)}^3=\frac{8}{27}$

${\left(\frac{3}{2}\right)}^{-1}={\left(\frac{2}{3}\right)}^1=\frac{2}{3}$

${\left(\frac{3}{2}\right)}^1=\frac{3}{2}$

${\left(\frac{2}{3}\right)}^{-2}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$

${\left(\frac{3}{2}\right)}^{-2}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$

${\left(\frac{2}{3}\right)}^0=1$

${\left(\frac{2}{3}\right)}^2=\frac{4}{9}$